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binary-tree-level-order-traversal solution (ts)
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/**
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* python의 len(queue)는 반복 시작 전에 계산된 값, 즉 고정된 횟수만큼 반복
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* JS의 queue.length는 반복마다 실시간으로 다시 계산하므로 고정 사이즈 size 변수 사용
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*/
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/**
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* Definition for a binary tree node.
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*/
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class TreeNode {
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val: number;
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left: TreeNode | null;
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right: TreeNode | null;
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constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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this.val = val === undefined ? 0 : val;
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this.left = left === undefined ? null : left;
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this.right = right === undefined ? null : right;
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}
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}
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function levelOrder(root: TreeNode | null): number[][] {
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if (!root) return [];
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const queue: TreeNode[] = [root];
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const output: number[][] = [];
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while (queue.length > 0) {
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const level: number[] = [];
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for (let node of queue) {
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level.push(node.val);
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}
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output.push(level);
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const size = queue.length;
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for (let i = 0; i < size; i++) {
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const node = queue.shift();
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if (node && node.left) {
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queue.push(node.left);
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}
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if (node && node.right) {
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queue.push(node.right);
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}
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}
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}
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return output;
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}

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